fengzhizi76506的博客

Big NumberTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 35518    Accepted Submission(s): 16950Problem DescriptionIn many applicati

七夕节Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 44674    Accepted Submission(s): 14241Problem Description七夕节那天,月老来到数字王国,他在城门上贴了一张

Assistance RequiredTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 2539    Accepted Submission(s): 1332Problem DescriptionAfter the

定义：欧拉函数f(n),表示小于或等于n的数中与n互质的数的个数通式：f(n)=n(1-1/p1)(1-1/p2)...(1-1/pn)其中p1, p2……pn为x的所有质因数，x是不为0的整数。φ(1)=1（唯一和1互质的数(小于等于1)就是1本身）。注意：每种质因数只一个。 比如12=2*2*3那么φ（12）=12*（1-1/2）*(1-1/3)=4求值方法

计算n的mobius函数u(n)定义：将n分解成素数幂的乘积：n=p1^i1*p2^i2*...*pr^iru(n) = { 1 (n==1)           0  (乘积式中至少有一个ir>1)          (-1)^r (i1==i2==...==ir==1)算法实现const int n = 1 int mu[n];int getMu(int n)

看了Simpson积分公式，惊讶于积分方法在几何问题中竟如此强大！比如几何中的求面积、体积问题，可以用积分的方法，在某个方向上用扫描线或扫描面切过图形，求被覆盖的长度或面积，然后进行积分。对于这类问题，如果能直接求出面积、体积，或者能列出覆盖长度（面积）关于扫描线（面）的位置的函数，然后手算积分，那当然是再好不过了。但是，如果直接用解析法、公式法比较费时费力，思考难

给定a,i,n,求a^i mod n每次将指数折半来计算。long long pow_mod(long long a,long long i,long long n){    if(i == 0) return 1%n;    int temp = pow_mod(a,i >> 1,n);        temp = temp*temp%n;    if(i&1) te

把一个x进制的数转换成y进制说明：把x进制的数转换成十进制，然后再不断取模再倒序，转换成y进制。2string transform(int x,int y,string s){ string res = ""; int sum = 0; for (int i = 0;i  {  if(s[i] >= '0' && s[i]   else sum =

给定一个二进制的位数n,求出一个0到2^n-1的排列，使得相邻两项（包含头尾两项）的二进制表达中只有一位恰好不同。说明：格雷码的第i位是i xor (i >> 1),简单的O(2^n)vector Grey_Create(int n){ vector res; res.clear(); for(int i = 0;i > 1)); return res;}

圆桌会议Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 4900    Accepted Submission(s): 3435Problem DescriptionHDU ACM集训队的队员在暑假集训时经常要讨论自

FactorialTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 3730    Accepted Submission(s): 2444Problem DescriptionThe most important p

Complete the SequenceTime Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 592    Accepted Submission(s): 388Problem DescriptionYou probab

Lifting the StoneTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 7634    Accepted Submission(s): 3234Problem DescriptionThere are ma

Factstone BenchmarkTime Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 2101    Accepted Submission(s): 1179Problem DescriptionAmtel has

Big NumberTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 7666    Accepted Submission(s): 5276Problem DescriptionAs we know, Big Num

CubeTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 1900    Accepted Submission(s): 1523Problem DescriptionCowl is good at solving m

The number of divisors(约数) about Humble NumbersTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3501    Accepted Submission(s): 1730Pro

find a way to escapeTime Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1958    Accepted Submission(s): 759Problem Description一日，话说0068与

CakeTime Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3750    Accepted Submission(s): 1962Problem Description一次生日Party可能有p人或者q人参加,现准备有

改革春风吹满地Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 30642    Accepted Submission(s): 15781Problem Description“ 改革春风吹满地,不会AC没关系;

num,den是分子，分母struct Fraction{ long long num; long long den; Fraction(long long num = 0,long long den = 1) {  if (den   {   num = -num;   den = -den;  }  assert(den != 0)//满足条件时报错